| Author | Message | ||
     Philip (riche) Member Username: riche Post Number: 5 Registered: 06-2006 |
Hello every one, I hope that someone know the same thing ''Transfermesh'' in Flex Pde 2.20d 3D. Two files below work very well with version 5.0.7 but not with version 2.20. I am seen in Help and find that there isn't command ''Transfermesh''in 2.20. But what is the same idea to import the mesh in another application in version 2.20? Thank you so much for your attention. Mai { MESH_OUT.PDE } { This example uses a modification of the sample problem HEATBDRY.PDE to illustrate the use of the TRANSFERMESH output function. Both the Temperatures calculated here and the final mesh structure are transferred as input to the stress calculation MESH_IN.PDE } title "Test TRANSFERMESH output" Variables Temp definitions K = 1 source = 4 Tzero = 0 flux = -K*grad(Temp) equations div(K*grad(Temp)) + source = 0 boundaries Region 1 start "OUTER" (0,0) natural(Temp)=0 line to(1,0) natural(Temp)=0 arc (center=0,0) to (0,1) natural(Temp)=0 line to close start "INNER" (0.4,0.2) natural(Temp)=Tzero-Temp arc (center=0.4,0.4) to (0.6,0.4) to (0.4,0.6) to (0.2,0.4) to close monitors contour(Temp) plots grid(x,y) contour(Temp) surface(Temp) vector(-K*dx(Temp),-K*dy(Temp)) as "Heat Flow" contour(source) elevation(normal(flux)) on "outer" range(-0.08,0.08) report(bintegral(normal(flux),"outer")) as "bintegral" elevation(normal(flux)) on "inner" range(1.95,2.3) report(bintegral(normal(flux),"inner")) as "bintegral" { HERE IS THE MESH TRANSFER OUTPUT COMMAND: } transfer(Temp) file="transferm.dat" { MESH_IN.PDE } { This problem demonstrates the use of the TRANSFERMESH facility to import both data and mesh structure from MESHOUT.PDE. MESH_OUT.PDE must be run before running this problem. } Title 'Testing the TRANSFERMESH statement' select painted { paint all contour plots } variables U V definitions nu = 0.3 { define Poisson's Ratio } E = 21 { Young's Modulus x 10**-11 } G = E/(1-nu**2) C11 = G C12 = G*nu C22 = G C33 = G*(1-nu)/2 alpha = 1e-3 b = G*alpha*(1+nu) { HERE IS THE TRANSFERMESH INPUT FUNCTION: } transfermesh('transferm.dat',Temp) Sxx = C11*dx(U) + C12*dy(V) - b*Temp Syy = C12*dx(U) + C22*dy(V) - b*temp Sxy = C33*(dy(U) + dx(V)) initial values U = 0 V = 0 equations U: dx[Sxx] + dy(Sxy) = 0 V: dy[Syy] + dx(Sxy) = 0 boundaries Region 1 start "OUTER" (0,0) natural(U)=0 value(V)=0 { no y-motion on x-axis } line to(1,0) natural(U)=0 natural(V)=0 { free outer boundary } arc (center=0,0) to (0,1) value(U)=0 natural(V)=0 { no x-motion on y-axis } line to close natural(U)=0 natural(V)=0 { free inner boundary } start "INNER" (0.4,0.2) arc (center=0.4,0.4) to (0.6,0.4) to (0.4,0.6) to (0.2,0.4) to close monitors grid(x+100*U,y+100*V) plots contour(Temp) grid(x+100*U,y+100*V) vector(U,V) as "Displacement" contour(U) as "X-Displacement" contour(V) as "Y-Displacement" contour(Sxx) as "X-Stress" contour(Syy) as "Y-Stress" surface(Sxx) as "X-Stress" surface(Syy) as "Y-Stress" | ||
| Philip (riche) Member Username: riche Post Number: 12 Registered: 06-2006 |
Dear Dr. Nelson I’d like to kwow the symmetrical axis in my probleme 2,5D when I make a gap (decalage, translation) r=d of my domaine in 0z direction to avoid 1/r = infini when r=0 : COORDINATES ycylinder('r','z') BOUNDARIES REGION 1 'Vacuum' nr=nr1 ni=ni1 Natural(Mtr)=k0*Mti Natural(Mti)=-k0*Mtr Start(d,-Rext) arc(center=d,0)angle=360 to finish REGION 2 'Matérial' nr=nr2 ni=ni2 Natural(Mtr)=0 Natural(Mti)=0 Start(d,-R0) arc(center=d,0)angle=360 to finish So, What is the the symmetrical axis of problem: r=0 or r=d ? Thank you | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 744 Registered: 06-2003 |
In cylindrical coordinates, the 2D figure is rotated 2*pi radians about the axis r=0. You have constructed a figure with both positive and negative radii, so that when this figure is rotated you will have doubly defined volumes. In order to use cylindrical coordinates correctly, you should have no negative radii, and bound your figure at the rotation axis r=0. If you bound the figure at r=d, you will have a tiny cylindrical hole in the center of your sphere. Usually this causes no trouble. But you must apply natural()=0 on this boundary. | ||
| Philip (riche) Member Username: riche Post Number: 13 Registered: 06-2006 |
Dear Dr, Does you mean that my rotation axis will be at r=d ( not at r=0) if my script is: COORDINATES ycylinder('r','z') BOUNDARIES REGION 1 'Vacuum' nr=nr1 ni=ni1 Natural(Mtr)=k0*Mti Natural(Mti)=-k0*Mtr Start(d,-Rext) arc(center=d,0)angle=180 Natural(Mtr)=0 Natural(Mti)=0 line to finish REGION 2 'Matérial' nr=nr2 ni=ni2 Natural(Mtr)=0 Natural(Mti)=0 Start(d,-R0) arc(center=d,0)angle=180 {R0<Rext} Natural(Mtr)=0 Natural(Mti)=0 line to finish And Finaly, I will have a sphere with her center at r=d ? Thank you | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 754 Registered: 06-2003 |
"Radius" is by definition the distance from the axis of rotation. Rotation is therefore always about r=0. The script you show will create a ring (torus) with a flat inner face at r=d and a semicircular outer face extending to r=d+Rext. If you want a sphere, the bounding line must be an arc centered at r=0. |